An Introduction To Differential Geometry (Part 4)

 Arc  length and Curvature:

 In our treatment of differential geometry,  arc length  is given by:

L = Öå ³ _i =1 ( y _i  -  x _i ) ²

Where ( x _i , y _i) define the end points of a curve segment. Hence, when defining the length of an arc C of a curve we can technically approximate it using a sequential series of broken lines or chords, as depicted below:

The length of such broken line can easily be determined if the end points of each chord are known. Let ℓ_i  then be a series of broken lines or chords whose end points lie on C.  Then if ℓ_i   tends to some limit s,  C is said to be rectifiable and is called the length of the arc C.  Let x(t) then (a <  t  <  b) be an allowable representation of an arc of a curve of class r > 1.  Then the arc C has length: 

s = ò ^b _a    Ö {å ³ _i =1  (dx _i /dt) ² } dt  

And s is independent of the choice of allowable representation. Note that if we replace the fixed value b in the above with the variable t then s becomes a function of t. Note also that a can be replaced with any other fixed value. Thereby we obtain the integral:

s(t) =  ò ^b _a     Ö(x' · x') dt 

The function s(t) is then called the arc length of C. If  t _o  > t then s(t) is positive and equal to the arc a(t _o) b (t)  of C. If t< t _o  then s(t) is negative and the length of a(t _o) b (t)  is given by - s(t). Instead of:

s² =  å ³ _i =1 x _i ²   =  x' · x'

We can write:  ds² = å ³ _i =1 dx _i ²   =  dx · dx

where  ds is called the element of arc.  For example, looking at the circular helix we find:  

x'  =  (- r sin t, r cos t, c),  x' · x'  =  r² + c ²

So that:  s(t)  =  tÖ r² + c ²

Then if we also wish to represent the circular helix by a parametric representation with arc length s as parameter, we can write:  

x(s) =  (r cos (s/w), r sin (s/w), cs/w )  And:

w=   Ö r² + c ²

What about?: 

x ₁ = t,   

x ₂ = t sin 1/t   {t ≠  0  

                            0  {t = 0

x ₃ = 0   (0 <  t   <  t ₁

We note the function is continuous even at t = 0, but the corresponding point set - which doesn't form an arc according to the definition, has no length.

Example (3): If the arc length in polar coordinates can be obtained from, e.g.:  

Find the arc length between  0  and  p/2,  then the arc length between  - p/2  and  p/2 for the Archimedean spiral:

 r  =  q - sin  q, and sketch the curve.  

Soln.  By integration from the polar eqn.

We will need to perform:

r(q) ²   = (q - sin  q) ²  

=  (q - sin  q) (q - sin  q) =

q ²  - 2 q sin  q +  sin ²  q

And:   d r(q)  / d q   =  1 -  cos  q

(d r(q)  / d q ) ²   = 

(1 - cos  q) (1 - cos  q)  =  1  - 2 cos  q  +  cos ²  q

Whence:

r(q) ²  +   (d r(q)  / d q ) ²   = 

q ²  - 2 q sin  q +  sin ²  q  +   1  - 2 cos  q  +  cos ²  q 

Leaving the integral:

ò ^{p / 2} ₀ Ö{( q ²  - 2 q sin  q +  sin ²  q)  + ( 1  -  2 cos  q  +  cos ²  q)} d q 

=>   ò ^{p / 2} ₀ Ö{q ²  -  2 cos  q - 2 q sin  q  +2 } d q

(Rem:  sin ² q  +  cos ²  q  =  1) 

From direct Mathcad computation:

And the full curve sketched :

The second integral is analogous - simply changing the limits - whereupon we find:

Suggested Problems:

1)A curve is given in spherical coordinates x _i  by:

x ₁ = t,    x ₂ = arcsin 1/t,    x ₃  =  (t ₂ – 1) ^1/2

 Compute the length of the arc between t = 1 and t = 2  

 2) Obtain the arc length s of the curve:

f(x) =  x³ / 2   -   x² / 3  

Between x= 0 and x=  2

3) Determine the arc length of a catenary with parametric representation:  x(t) =  (t,  a cosh (t/a), 0)  

4) Find the full arc length of the Archimedian spiral shown by changing the integral to the correct limits.